Solve frac{5+sqrt{3}i}{2-sqrt{3}i}

$$\frac { 5 + \sqrt { 3 } i } { 2 - \sqrt { 3 } i }$$

Evaluate

$$1+\sqrt{3}i\approx 1+1.732050808i$$

Solution Steps

Rationalize the denominator of \frac{5+i\sqrt{3}}{2-i\sqrt{3}} by multiplying numerator and denominator by 2+i\sqrt{3}.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{\left(2-i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}$$

Consider \left(2-i\sqrt{3}\right)\left(2+i\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{2^{2}-\left(-i\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{4-\left(-i\sqrt{3}\right)^{2}}$$

Expand \left(-i\sqrt{3}\right)^{2}.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{4-\left(-i\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -i to the power of 2 and get -1.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{4-\left(-\left(\sqrt{3}\right)^{2}\right)}$$

The square of \sqrt{3} is 3.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{4-\left(-3\right)}$$

Multiply -1 and -3 to get 3.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{4+3}$$

Add 4 and 3 to get 7.

$$\frac{\left(5+i\sqrt{3}\right)\left(2+i\sqrt{3}\right)}{7}$$

Apply the distributive property by multiplying each term of 5+i\sqrt{3} by each term of 2+i\sqrt{3}.

$$\frac{10+5i\sqrt{3}+2i\sqrt{3}-\left(\sqrt{3}\right)^{2}}{7}$$

Combine 5i\sqrt{3} and 2i\sqrt{3} to get 7i\sqrt{3}.

$$\frac{10+7i\sqrt{3}-\left(\sqrt{3}\right)^{2}}{7}$$

The square of \sqrt{3} is 3.

$$\frac{10+7i\sqrt{3}-3}{7}$$

Subtract 3 from 10 to get 7.

$$\frac{7+7i\sqrt{3}}{7}$$

Real Part

$$1$$